1。 Because f'(x)=2x-2a, and because f(x) decreases on (negative infinity, 2). So f' (x)

Because a & gt=2, a+1>; So the minimum point of the function is on x=a, and the maximum value can only be f( 1) or f(a+ 1), so there is

| f( 1)-f(a)| & lt； =4，| f(a+ 1)-f(a)| & lt； =4 amps

2.( 1) Because if a >: =0 at this time, it is obvious that h(x) is increasing function in this area. So h( 1)=3, so 0-a=3 leads to the contradiction of a=-3, so a

so-a & gt； =e，a & lt=-e，h(e)=3，a =-2e

(2) let h (x) = f (x)-g (x)-x 2, as long as x0 >；; = 1 makes h(x0)>0, and because h' (x) =1/x-2x+a/(x 2), when a